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Integration by parts is an integration technique that involves separating the integrand into two parts. This technique is useful whenever a product appears in the integrand and the implement u-substitution is not feasible.
\(\displaystyle\int u\,dv = uv\,-\!\int v\,du\)
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Answer the following questions by using the integration technique known as integration by parts (IBP).
Let \(u=x\), so \(du=dx\); and \(dv=\texttt{sin}(2x)dx\), so \(v=\int\texttt{sin}(2x)dx=-\frac{1}{2}\texttt{cos}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Let \(u=x^2\), so \(du=2xdx\); and \(dv=\texttt{sin}(2x)dx\), so \(v = \int \texttt{sin}(2x)dx = -\frac{1}{2}\texttt{cos}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Apply IBP once again. Let \(u=x\), so \(du=dx\); and \(dv=\texttt{cos}(2x)dx\), so \(v = \int \texttt{cos}(2x)dx = \frac{1}{2}\texttt{sin}(2x)\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Let \(u=\texttt{arctan}(2x+1)\), so \(du=\frac{2}{(2x+1)^2+1}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Use U-Sub with the given integrand. Let \(u=2x+1\), so \(du=2dx\); and \(x=\frac{u-1}{2}\). While working the only integral left, it’s fine to have \(u\) and \(x\) at the same time given that the integrand is expressed in regards of just one variable. Substituting accordingly yields:
\(\normalsize
Use U-Sub with the given integrand. Let \(v=u^2\), so \(dv=2udu\). Substituting accordingly yields:
\(\normalsize
Let \(u=\texttt{cos}(\sqrt{x})\), so \(du=\frac{-\texttt{sin}(\sqrt{x})}{2\sqrt{x}}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv – \int vdu\) leads to the following:
\(\normalsize
Use U-Sub with the given integrand. Let \(u=\sqrt{x}\), so \(du=\frac{1}{2\sqrt{x}}dx\); \(dx=2\sqrt{x}du=2udu\). Substituting accordingly yields:
\(\normalsize
Apply IBP again. Let \(m=u^2\), so \(dm=2udu\); and \(dw=\texttt{sin}(u)du\), so \(w=-\texttt{cos}(u)\). Substituting these values into \(mw\:- \int wdm\) leads to the following:
Apply IBP again. Let \(m=u\), so \(dm=du\); and \(dw=\texttt{cos}(u)du\), so \(w=\texttt{sin}(u)\). Substituting these values into \(mw\:- \int wdm\) leads to the following:
\(\normalsizeLet \(u=\texttt{e}^{\sqrt{x}}\), so \(du=\frac{\texttt{e}^{\sqrt{x}}}{2\sqrt{x}}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Use U-Sub with the given integrand. Let \(u=\sqrt{x}\), so \(du=\frac{1}{2\sqrt{x}}dx\); \(dx=2\sqrt{x}du=2udu\). Substituting accordingly yields:
\(\normalsize
Apply IBP. Let \(m=u^2\), so \(dm=2udu\); and \(dw=\texttt{e}^udu\), so \(w=\texttt{e}^u\). Substituting these values into \(mw\:- \int wdm\) leads to the following:
Apply IBP one last time. Let \(m=u\), so \(dm=du\); and \(dw=e^udu\), so \(w=e^u\). Substituting these values into \(mw\:- \int wdm\) leads to the following:
\(\normalsize
Let \(u=\texttt{sin(ln(x))}\), so \(du=\frac{\texttt{cos(ln(x))}}{x}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Use IBP again. Let \(u=\texttt{cos(ln(x))}\), so \(du=\frac{-\texttt{sin(ln(x))}}{x}dx\); and \(dv=dx\), so \(v=x\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:
\(\normalsizeLet \(u=\texttt{cos}(5x)\), so \(du=-5\texttt{sin}(5x)dx\); and \(dv=\texttt{e}^{2x}dx\), so \(v=\frac{1}{2}\texttt{e}^{2x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Use IBP again. Let \(u=\texttt{sin}(5x)\), so \(du=5\texttt{cos}(5x)dx\); and \(dv=\texttt{e}^{2x}dx\), so \(v=\frac{1}{2}\texttt{e}^{2x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:
\(\normalsizeLet \(u=\texttt{sin}(\beta x)\), so \(du=\beta \texttt{cos}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha}\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Use IBP again. Let \(u=\texttt{cos}(\beta x)\), so \(du=-\beta \texttt{sin}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha }\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsize
Given that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:
\(\normalsizeLet \(u=\texttt{cos}(\beta x)\), so \(du=-\beta \texttt{sin}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha}\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsizeUse IBP again. Let \(u=\texttt{sin}(\beta x)\), so \(du=\beta \texttt{cos}(\beta x)dx\); and \(dv=\texttt{e}^{\alpha x}dx\), so \(v=\frac{1}{\alpha }\texttt{e}^{\alpha x}\). Substituting these values into \(uv\:- \int vdu\) leads to the following:
\(\normalsizeGiven that we ended up with the same integral we wanted to solve for, we can equate the original integrand to the resulting expression and perform algebraic manipulations while treating the reappearing integral as a variable to solve for:
\(\normalsizeCheck with your tutor
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